Specifically I need to decompose $\frac1{(1-x)(1-x^n)^2}$ into $\frac{f(x)}{(1-x)^3}+\frac{g(x)}{1-x^n\vphantom{()^2}}+\frac{h(x)}{(1-x^n)^2}$ where $f(x)$, $g(x)$, $h(x)$ are polynomials.
Surely there must be well known methods to deal with this. Can I avoid taking $f$, $g$, $h$ with indeterminate coefficients and then solve a system of linear equations?
In one of the answers here I found a link to something called Heaviside cover-up method but there only denominators with terms up to degree 2 were dealt with.
Note that you want to decompose $\frac{1}{a(x)^3 b(x)^2}$, where $a(x)=1-x$ and $(1-x)b(x)=1-x^n$. But $a(x)^3$ and $b(x)^2$ are coprime polynomials, i.e. their greatest common divisor is 1. Hence there exist polynomials $f(x),h(x)$ such that $1=f(x) b(x)^2+h_1(x) a(x)^3$. Hence $$ \frac{1}{a(x)^3 b(x)^2}=\frac{f(x)}{a(x)^3}+\frac{h_1(x)}{b(x)^2}. $$ Set $h(x):=(1-x)^2 h_1(x)$ and then $$ \frac{1}{a(x)^3 b(x)^2}=\frac{f(x)}{(1-x)^3}+\frac{h(x)}{(1-x)^2 b(x)^2}= \frac{f(x)}{(1-x)^3}+\frac{h(x)}{(1-x^n)^2}. $$ Set $g(x)=0$.
You can find this in any abstract algebra book, or for example in section 5 of http://www.imsc.res.in/~knr/past/part_frac.pdf